Related Rates: Question 3-11.1 & Solution

Question 1

A streetlight is mounted at the top of a 20-foot pole. A person 6 feet tall walks directly away from the pole at a constant rate of 4 ft/sec.

(a) How fast is the tip of the person’s shadow moving along the ground when the person is 10 feet from the pole?

(b) How fast is the length of the person’s shadow changing at that instant?

(c) Explain why the tip of the shadow moves faster than the person.

See the diagram in the original worksheet below.

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Let: $x(t) = \text{distance from the pole to the person}$ $s(t) = \text{length of the shadow}$

The distance from the pole to the tip of the shadow is: $x + s$

Using similar triangles: $\frac{20}{x + s} = \frac{6}{s}$

Cross-multiply: $20s = 6(x + s)$

Simplify: $20s = 6x + 6s \Rightarrow 14s = 6x \Rightarrow s = \frac{3}{7}x$

Differentiate both sides: $\frac{ds}{dt} = \frac{3}{7} \frac{dx}{dt}$

Given: $\frac{dx}{dt} = 4$

So: $\frac{ds}{dt} = \frac{12}{7}$

$\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt} = 4 + \frac{12}{7} = \frac{40}{7}$

Answer (a): $\boxed{\frac{40}{7} \text{ ft/sec}}$

$\boxed{\frac{ds}{dt} = \frac{12}{7} \text{ ft/sec}}$

The shadow tip moves faster than the person because its motion includes both the person’s movement and the changing length of the shadow.

Conclusion: The shadow tip moves faster than the person.